Question: Balance the following chemical equation: $ $ $\text{H}_2\text{SO}_4 +$ $\text{Pb(OH)}_4 \rightarrow$ $\text{Pb(SO}_4\text{)}_2 +$ $\text{H}_2\text{O}$
Answer: We can treat the sulfate polyatomic ion $\text{(SO}_4\text{)}$ as an atom, symbolized by ${X}$ $ \text{H}_2{\text{SO}_4} + \text{Pb(OH)}_4 \rightarrow \text{Pb(}{\text{SO}_4}\text{)}_2 + \text{H}_2\text{O} $ $ \text{H}_2{X} + \text{Pb(OH)}_4 \rightarrow \text{Pb}{X}_2 + \text{H}_2\text{O} $ There is $1 \space X$ on the left and $2 \space X$ on the right, so multiply $\text{H}_2X$ by ${2}$ $ {2}\text{H}_2X + \text{Pb(OH)}_4 \rightarrow \text{Pb}X_2 + \text{H}_2\text{O} $ That gives us $8 \text{ H}$ on the left and only $2$ on the right, so multiply $\text{H}_2\text{O}$ by ${4}$ $ 2\text{H}_2X + \text{Pb(OH)}_4 \rightarrow \text{Pb}X_2 + {4}\text{H}_2\text{O} $ Everything is now balanced. Replacing $\text{SO}_4$ for $X$, the balanced equation is: $ 2\text{H}_2\text{SO}_4 + \text{Pb(OH)}_4 \rightarrow \text{Pb(}\text{SO}_4\text{)}_2 + 4\text{H}_2\text{O} $